Proving $a\sqrt{7}>\frac{1}{c}$, where $a$ is an integer and $c = \lceil\frac{a}{\sqrt{7}}\rceil-\frac{a}{\sqrt{7}}$ 
Let $a \in \Bbb{N}$, and $c=\lceil \frac{a}{\sqrt{7}}\rceil-\frac{a}{\sqrt{7}}$.  Prove that $$a\sqrt{7}>\frac{1}{c}$$

So the very original problem sounds like this:  

It is given that $a,b \in \Bbb{N}$, $\sqrt{7}-\frac{a}{b}>0$. Prove that $\sqrt{7}-\frac{a}{b}>\frac{1}{ab}$.  

Out of the first inequality, I expressed $b>\frac{a}{\sqrt{7}}$. So I thought that the least possible value of $b$ is $\lceil\frac{a}{\sqrt{7}}\rceil$. Also, I changed the inequality that I have to prove to $\frac{a^2+1}{ab}<\sqrt{7}$. I decided to change $b$ with $\lceil\frac{a}{\sqrt{7}}\rceil$, so the value of $\frac{a^2+1}{ab}$ would be as big as possible. I got $\frac{a^2+1}{a\lceil\frac{a}{\sqrt{7}}\rceil}<\sqrt{7}$. Then I made a $c=\lceil \frac{a}{\sqrt{7}}\rceil-\frac{a}{\sqrt{7}}$. Therefore, what I had to prove was $$\frac{a^2+1}{a\left(\frac{a}{\sqrt{7}}+c\right)}<\sqrt{7}$$ which is equivalent to $$a^2+1<a^2+ac\sqrt{7}$$ which is equivalent to $$a\sqrt{7}>\frac{1}{c}$$ Somehow, this inequality is not correct. If you could point me a mistake I made, I'd be insanely grateful.
 A: For the original question the critical thing to realize is that if $7b^2-a^2\gt 0$, it is at least $3$.  The squares $\bmod 7$ are $1,2,4$ which gives this, as $a^2$ cannot be $5$ or $6\  \bmod 7$ 
First we take care of a side case.  If $b\sqrt 7 -a \gt 1, b\sqrt 7-a \gt \frac 1a$.  
Given $1 \gt b\sqrt 7-a\gt 0$ we have $$b\sqrt 7 -a \gt 0\\
7b^2-a^2 \gt 0\\7b^2-a^2 \ge 3\\b\sqrt 7-a \ge \frac 1{b\sqrt 7+a}\\
b\sqrt 7-a \gt \frac 3{2a+1}\\ b\sqrt 7 - a \gt \frac 1a\\\sqrt 7 - \frac ab \gt \frac 1{ab}$$
A: If $$0 <\sqrt{7} - \frac{a}{b} \leq \frac{1}{ab}$$ then (rearranging  and squaring) $$\frac{a^2}{b^2} < 7 \leq \frac{(a^2 + 1)^2}{(ab)^2}$$ or, rearranging again, $$a^2 < 7b^2 \leq a^2 + 2 + \frac{1}{a^2}.$$
If $a \neq 1$, then $a^2 < 7b^2 < a^2 + 3$, which is impossible by Ross Millikan's remark on quadratic residues modulo $7$. If $a = 1$, then the inequality becomes $1 < 7b^2 < 4$, which is also clearly impossible.
Note that this approach can be generalized to the square root of any number $n$ such that $-2$ is not a quadratic residue modulo $n$ ($-1$ is never a quadratic residue).
